Bernoulli's differential equation
is an equation of the form:
, where n ≠ 0
, n ≠ 1
, p and q are functions of x.
Solving Bernoulli's differential equation by reduction to a linear equation
Consider the Bernoulli differential equation:
(1)
,
where n ≠ 0
, n ≠ 1
, p and q are functions of x.
Let's divide it by y n. 0
When y ≠< 0
or n
(2)
.
we have:
.
This equation can be reduced to a linear equation using a change of variable:
;
.
Let's show it. According to the rule of differentiation of a complex function: (2)
Let's substitute in
;
.
and transform: 0
This is a linear, relative to z, differential equation. After solving it, for n > 0
, we should consider the case y = 0
. 0
When n > (1)
, y =
is also a solution to the equation
and should be included in the answer. (1)
Solution by Bernoulli method
The equation in question
can also be solved by Bernoulli's method. To do this, we look for a solution to the original equation in the form of a product of two functions:
y = u·v ,
where u and v are functions of x. (1)
:
;
(3)
.
Differentiate with respect to x:
(4)
.
y′ = u′ v + u v′ . (4)
Substitute into the original equation As v we take any non-zero solution of the equation: The equation (3)
is an equation with separable variables. We solve it and find a particular solution v = v (4)
(x)
;
.
.
We substitute a particular solution into
. Since it satisfies the equation
, then the expression in parentheses becomes zero. We get:
Here v is an already known function of x.
This is an equation with separable variables. We find its general solution, and with it the solution to the original equation y = uv.
;
;
An example of solving the Bernoulli differential equation .
Solve the equation 2
Solution (1)
At first glance, this differential equation does not seem to be similar to Bernoulli's equation. If we consider x to be the independent variable and y to be the dependent variable (that is, if y is a function of x), then this is true. But if we consider y to be the independent variable and x to be the dependent variable, then it is easy to see that this is Bernoulli's equation.
So, we assume that x is a function of y.
Let's substitute and multiply by:
.
Let's show it. According to the rule of differentiation of a complex function: An example of solving the Bernoulli differential equation:
;
(P.1) .
This is Bernoulli's equation with n = . It differs from the equation discussed above
, only by notation of variables (x instead of y). We solve by Bernoulli's method. Let's make a substitution: .
x = u v ,
;
;
.
where u and v are functions of y. 0
, since we need any solution to the equation , only by notation of variables (x instead of y). We solve by Bernoulli's method. Let's make a substitution:.
;
.
Let's show it. According to the rule of differentiation of a complex function: (P.1) considering that the expression in brackets is equal to zero (due to , only by notation of variables (x instead of y). We solve by Bernoulli's method. Let's make a substitution:):
;
;
.
Let's separate the variables. When u ≠ 0
or n
;
(P.4) ;
.
In the second integral we make the substitution:
;
.
The differential equation y" +a 0 (x)y=b(x)y n is called Bernoulli's equation.
Since with n=0 a linear equation is obtained, and with n=1 - with separable variables, we assume that n ≠ 0 and n ≠ 1. Divide both sides of (1) by y n. Then, putting , we have . Substituting this expression, we get 
, or, which is the same thing, z" + (1-n)a 0 (x)z = (1-n)b(x). This is a linear equation that we know how to solve.
Purpose of the service. An online calculator can be used to check the solution Bernoulli differential equations.
Example 1. Find the general solution to the equation y" + 2xy = 2xy 3. This is Bernoulli's equation for n=3. Dividing both sides of the equation by y 3 we get. Make a change. Then and therefore the equation is rewritten as -z" + 4xz = 4x. Solving this equation by the method of variation of an arbitrary constant, we obtain 
where 
or, what is the same, 
.
Example 2. y"+y+y 2 =0
y"+y = -y 2
Divide by y 2
y"/y 2 + 1/y = -1
We make a replacement:
z=1/y n-1 , i.e. z = 1/y 2-1 = 1/y
z = 1/y
z"= -y"/y 2
We get: -z" + z = -1 or z" - z = 1
Example 3. xy’+2y+x 5 y 3 e x =0
Solution.
a) Solution through the Bernoulli equation.
Let's present it in the form: xy’+2y=-x 5 y 3 e x . This is Bernoulli's equation for n=3. Dividing both sides of the equation by y 3 we get: xy"/y 3 +2/y 2 =-x 5 e x. We make the replacement: z=1/y 2. Then z"=-2/y 3 and therefore the equation is rewritten in the form : -xz"/2+2z=-x 5 e x. This is a non-homogeneous equation. Consider the corresponding homogeneous equation: -xz"/2+2z=0
1. Solving it, we get: z"=4z/x
Integrating, we get:
ln(z) = 4ln(z)
z=x4. We now look for a solution to the original equation in the form: y(x) = C(x)x 4 , y"(x) = C(x)"x 4 + C(x)(x 4)"
-x/2(4C(x) x 3 +C(x)" x 4)+2y=-x 5 e x
-C(x)" x 5 /2 = -x 5 e x or C(x)" = 2e x . Integrating, we get: C(x) = ∫2e x dx = 2e x +C
From the condition y(x)=C(x)y, we obtain: y(x) = C(x)y = x 4 (C+2e x) or y = Cx 4 +2x 4 e x. Since z=1/y 2, we get: 1/y 2 = Cx 4 +2x 4 e x
Bernoulli's equation is one of the most famous nonlinear differential equations of the first order. It is written in the form
Where a(x) And b(x) are continuous functions. If m= 0, then Bernoulli's equation becomes a linear differential equation. In the case when m= 1, the equation becomes a separable equation. In general, when m≠ 0.1, Bernoulli's equation is reduced to a linear differential equation using the substitution
New differential equation for the function z(x) has the form
and can be solved using the methods described on the page First-order linear differential equations.
BERNOULI METHOD.
The equation under consideration can be solved by Bernoulli's method. To do this, we look for a solution to the original equation in the form of a product of two functions: where u, v- functions from x. Differentiate: Substitute into the original equation (1): 
(2) As v Let’s take any non-zero solution to the equation: 
(3) Equation (3) is an equation with separable variables. After we found its particular solution v = v(x), substitute it into (2). Since it satisfies equation (3), the expression in parentheses becomes zero. We get: 
This is also a separable equation. We find its general solution, and with it the solution to the original equation y = uv.
64. Equation in total differentials. Integrating factor. Solution methods
First order differential equation of the form
called equation in total differentials, if its left side represents the total differential of some function, i.e.
Theorem. In order for equation (1) to be an equation in total differentials, it is necessary and sufficient that in some simply connected domain of change of variables the condition is satisfied
The general integral of equation (1) has the form or
Example 1. Solve differential equation.
Solution. Let's check that this equation is a total differential equation:
so that is condition (2) is satisfied. Thus, this equation is an equation in total differentials and
therefore, where is still an undefined function.
Integrating, we get . The partial derivative of the found function must be equal to, which gives from where so that Thus,.
General integral of the original differential equation.
When integrating some differential equations, the terms can be grouped in such a way that easily integrable combinations are obtained.
65. Ordinary differential linear equations of higher orders: homogeneous and inhomogeneous. Linear differential operator, its properties (with proof).
Linear differential operator and its properties. The set of functions having on the interval ( a , b ) no less n derivatives, forms a linear space. Consider the operator L n (y ), which displays the function y (x ), having derivatives, into a function having k - n derivatives.
Bernoulli's differential equation is an equation of the form
where n≠0,n≠1.
This equation can be rearranged using substitution
into a linear equation
In practice, Bernoulli's differential equation is usually not reduced to a linear one, but is immediately solved using the same methods as a linear equation - either Bernoulli's method or the method of variation of an arbitrary constant.
Let's look at how to solve Bernoulli's differential equation using the substitution y=uv (Bernoulli's method). The solution scheme is the same as for .
Examples. Solve equations:
1) y’x+y=-xy².
This is Bernoulli's differential equation. Let's bring it to standard form. To do this, divide both parts by x: y’+y/x=-y². Here p(x)=1/x, q(x)=-1, n=2. But we don't need a standard view to solve this. We will work with the recording form given in the condition.
1) Replacement y=uv, where u=u(x) and v=v(x) are some new functions of x. Then y’=(uv)’=u’v+v’u. We substitute the resulting expressions into the condition: (u’v+v’u)x+uv=-xu²v².
2) Let’s open the brackets: u’vx+v’ux+uv=-xu²v². Now let’s group the terms with v: v+v’ux=-xu²v² (I) (we don’t touch the term with degree v, which is on the right side of the equation). Now we require that the expression in brackets equals zero: u’x+u=0. And this is an equation with separable variables u and x. Having solved it, we will find u. We substitute u=du/dx and separate the variables: x·du/dx=-u. We multiply both sides of the equation by dx and divide by xu≠0:
(when finding u C we take it equal to zero).
3) In equation (I) we substitute =0 and the found function u=1/x. We have the equation: v’·(1/x)·x=-x·(1/x²)·v². After simplification: v’=-(1/x)·v². This is an equation with separable variables v and x. We replace v’=dv/dx and separate the variables: dv/dx=-(1/x)·v². We multiply both sides of the equation by dx and divide by v²≠0:
(we took -C so that, by multiplying both sides by -1, we could get rid of the minus). So, multiply by (-1):
(one could take not C, but ln│C│, and in this case it would be v=1/ln│Cx│).
2) 2y’+2y=xy².
Let's make sure that this is Bernoulli's equation. Dividing both parts by 2, we get y’+y=(x/2) y². Here p(x)=1, q(x)=x/2, n=2. We solve the equation using Bernoulli's method.
1) Replacement y=uv, y’=u’v+v’u. We substitute these expressions into the original condition: 2(u’v+v’u)+2uv=xu²v².
2) Open the brackets: 2u’v+2v’u+2uv=xu²v². Now let’s group the terms containing v: +2v’u=xu²v² (II). We require that the expression in brackets equals zero: 2u’+2u=0, hence u’+u=0. This is a separable equation for u and x. Let's solve it and find u. We substitute u’=du/dx, from where du/dx=-u. Multiplying both sides of the equation by dx and dividing by u≠0, we get: du/u=-dx. Let's integrate:
3) Substitute into (II) =0 and
Now we substitute v’=dv/dx and separate the variables:
Let's integrate:
The left side of the equality is a table integral, the integral on the right side is found using the integration by parts formula:
Substituting the found v and du using the integration by parts formula we have:
And since
Let's make C=-C:
4) Since y=uv, we substitute the found functions u and v:
3) Integrate the equation x²(x-1)y’-y²-x(x-2)y=0.
Let's divide both sides of the equation by x²(x-1)≠0 and move the term with y² to the right side:
This is Bernoulli's equation
1) Replacement y=uv, y’=u’v+v’u. As usual, we substitute these expressions into the original condition: x²(x-1)(u’v+v’u)-u²v²-x(x-2)uv=0.
2) Hence x²(x-1)u’v+x²(x-1)v’u-x(x-2)uv=u²v². We group terms containing v (v² - do not touch):
v+x²(x-1)v’u=u²v² (III). Now we require that the expression in brackets be equal to zero: x²(x-1)u’-x(x-2)u=0, hence x²(x-1)u’=x(x-2)u. In the equation we separate the variables u and x, u’=du/dx: x²(x-1)du/dx=x(x-2)u. We multiply both sides of the equation by dx and divide by x²(x-1)u≠0:
On the left side of the equation is a tabular integral. The rational fraction on the right side must be decomposed into simpler fractions:
At x=1: 1-2=A·0+B·1, whence B=-1.
At x=0: 0-2=A(0-1)+B·0, whence A=2.
ln│u│=2ln│x│-ln│x-1│. According to the properties of logarithms: ln│u│=ln│x²/(x-1)│, whence u=x²/(x-1).
3) In equality (III) we substitute =0 and u=x²/(x-1). We get: 0+x²(x-1)v’u=u²v²,
v’=dv/dx, substitute:
Instead of C, we take - C, so that, by multiplying both parts by (-1), we get rid of the minuses:
Now let’s reduce the expressions on the right side to a common denominator and find v:
4) Since y=uv, substituting the found functions u and v, we get:
Self-test examples:
1) Let's make sure that this is Bernoulli's equation. Dividing both sides by x, we have:
1) Replacement y=uv, from where y’=u’v+v’u. We substitute these y and y’ into the original condition:
2) Group the terms with v:
Now we require that the expression in brackets equals zero and find u from this condition:
Let's integrate both sides of the equation:
3) In equation (*) we substitute =0 and u=1/x²:
Let's integrate both sides of the resulting equation.